text:
Do you pitch the nose up or down when flying at best glide speed during a glide
So in a glide, flying at the best glide speed, will you be pitching the nose up (nose points above the horizon) or pitching the nose down (nose points below the horizon) or does it depend?

In practice you will fly rarely with the best glide speed. In thermics you will fly with the speed of lest decend or a little below to stay as long as possible in the upwind. Between thermics you will fly faster then the best glide speed since between thermics the air descends an you want to pass that fast. How much you exceed the best gliding speed depends on the strength of the thermics that in turn determines the strength of the descending air inbetween... see: https://en.wikipedia.org/wiki/Speed_to_fly

Are you asking about the initial change in the nose attitude required to establish steady state flight at whatever best glide speed is? or the nose attitude required to maintain that steady state speed once established? In either case, there are many factors involved. The aircraft itself being the biggest one.

You pitch at whatever angle and attitude is required to maintain your desired speed. That's why it's called best glide speed, not best glide angle.
Having said that, unless you were already descending, it's going to end up with the nose being at or below the horizon since, well, you're gliding...

Not so, in a high performance swept wing jet, (most fighter aircraft, or Concorde, for example), the AOA at best glide speed is greater than the flight path glide angle, and the nose will be above the horizon...

@CharlesBretana Right, and what do you think the chances are that the OP is flying any of those types?

The example given (high performance swept wing, is just the extreme. The point made ( that AOA and glide angle are the determinants), does not only apply there.  Using another more common example, many GA aircraft in a no-flap configuration, have the nose above the horizon in a glide.

Your comment is wrong because you conflate nose or pitch attitude with flight path angle. Just because, as you comment ("since, well, you're gliding..."), doesn't mean that the nose pitch attitude has to be below the horizon.   Using yet another (less common) example, in extremely high performance gliders, for example, where the glide ratios are very high (some are 70:1), even a very small AOA will put the nose above the horizon.

Surely the key point is not the nose, but the wing itself?

@MikeB,  Not sure what your point is, but the op's question asks about the "... nose up or down...", not the wing

Indeed, in another example, the Navy F-8 Crusader had the capability to increase the angle of incidence of the wing to increase pilot visibility over the nose during landings. Making the angle of the wing to the fuselage higher lowered the nose attitude at the same glide path angle.

Are you asking about the initial change in the nose attitude required to establish steady state flight at whatever best glide speed is? or the nose attitude required to maintain that steady state speed once established? In either case, there are many factors involved. The aircraft itself being the biggest one.
If you are asking about the initial pitch required, then there are so many variables, that there is no clear answer....
But if you are asking about the steady state situation, once you have established it, then in general, the answer depends on the difference between the flight path glide angle, at best glide speed, and the Angle of Attack (AOA) required to hold that glide angle. If the AOA is greater than the glide angle, then the nose will be above the horizon, if it is less....


text:
Altitude/Minimum altitude in ARINC 424
Airport SID/STAR/Approach Primary Records in ARINC 424 contains two fields with altitudes: columns 85-89 and columns 90-94. How to interpret if both fields are empty ? What are the altitude constraints in this case?
Thanks in advance!

Could it be that there is no hard altitude at these fixes?  If you provide the actual SID/Star this data comes from I'd be happy to check.

Actually, I don't have any charts that cover Moscow...  But, if you post the published procedure it might still help.

@MichaelHall is correct. A blank altitude block indicates no altitude restriction. If you have a copy of the actual ARINC 424 manual, it explains what values may be entered in these blocks as well as the preceding column that indicates a + (minimum), - (maximum), or B (block) restriction. This has changed quite a bit since 424-18. There used to be a lot more available options.

Thanks for your answers! I have ARINC manual. The manual have description of field 83 (Altitude description) and blank in this field means "“At” altitude specified in first “Altitude” field.". But what if the first "Altitude" field is empty? Let's assume that it means no altitude restrictions. Does this mean that the aircraft can fly at an altitude from the lower boundary of the sector to the upper one?

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text:
Weird reroutes for two flights flying out of KJFK
A few months ago, I was tracking some flights on Flightaware, and on July 11, 2024, I was quite surprised to see these two flights from KJFK to RKSI (Incheon, South Korea) depart toward the south, fly past New Jersey and Delaware, and then finally turn northbound while over Maryland.
KAL86:

https://www.flightaware.com/live/flight/KAL86/history/20240711/0500Z/KJFK/RKSI
AAR223:

https://www.flightaware.com/live/flight/AAR223/history/20240711/0445Z/KJFK/RKSI
I tried listening to LiveATC's archived audio, and it seems that AAR223 (and presumably KAL86) was rerouted so that its initial waypoint was WAVEY (I just heard the departure controller instruct AAR223 to proceed direct to WAVEY. The reroute is not reflected in the route displayed on Flightaware for some reason).
My question is: does anyone know if there was anything specific going on at JFK in the early mornings of July 11, 2024 that prevented aircraft from using the East or North departure gates? I fly this route often; we've always departed toward the north to GAYEL or east toward MERIT or GREKI. WAVEY (and then flying over New Jersey, Delaware, and Maryland) seems like a weird reroute for a flight going to East Asia, and I'd love to get some information on why this happened. This happened with two different airlines on the same day around the same time frame, so I would think this would be on the part of the ATC and not the airlines' dispatchers.
Interestingly, this was not the case with ANA159, which departed for RJTT (Ota, Japan) on that same day in the early morning. It seems to have turned north toward GAYEL while still in N90's airspace.
Bonus points if anyone manages to get the actual routing that those flights used.
ANA159:

https://www.flightaware.com/live/flight/ANA159/history/20240711/0610Z/KJFK/RJTT

Those pilots just really hate Jersey and Yonkers

TL;DR: Weather.
I went to the FAA's ATC System Command Center Advisories page to see what advisories were issued on that day. The dropdown menu only goes back a couple weeks but once I was on the actual list/database page it was simple to edit the URL to reference "Thursday%2C+07-11-2024" instead of "Friday%2C+09-20-2024." On that page I found Advisory 001, NE_DELMARVA_ESCAPE, which was actually a continuation of the previous day's Advisory 113.
The escape routes were due to weather, as referenced in the 10/2200Z ops plan. Thunderstorms were affecting the normal West departure routes out of the New York area.
Per the advisory, the required routing to "escape" JFK was WAVEY EMJAY Q167 ZJAAY Q97 SAWED. Changing the FlightAware layer to the enroute sectional we can see that AAR223 didn't go all the way on that route but instead got a shortcut from before ZJAAY to either PXT VORTAC or HYTRA waypoint, and from there went direct SAW and then joined the route shown by FlightAware. KAL86 got their shortcut a little sooner, around TBONN.

It is interesting that the advisory says it's valid for aircraft with expected departure times up until 11/0230Z, but the two aircraft in question departed over two hours after that time and still got the reroute. Perhaps they had already fueled up expecting the longer route? Or perhaps the weather ended up being a problem for longer than expected? I'm not sure.

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text:
FAA - Common Aircraft Categories
I understand that for Small Airplanes, the FAA establishes different categories, one of which is Normal, Utility, and Aerobatic, commonly known for their Limit Load Factors, which are:

Normal: 3.8 G, -1.52G

Utility: 4.4 G, -1.76G

Aerobatic: 6.0 G, -3.0 G

It's worth noting that these are categories for light aircraft, meaning those with a certified Maximum Takeoff Weight (MTOW) of 12,500 pounds or less. However, I have not found any official information from the FAA that takes these limits into account for the certification of the Normal category, only the MTOW. Upon further investigation in 14 CFR, I found that in Section 23.2005, it states that one of the requirements for the Certification of normal category airplanes is: "Certification in the normal category applies to airplanes with a passenger-seating configuration of 19 or less and a maximum certificated takeoff weight of 19,000 pounds or less." It also refers to which aerobatic aircraft may or may not be certified in the normal category, as can be seen in part (d), which is also weird because they are "different categories"
This appears to contradict the previously mentioned Normal Category, which allows aircraft of up to 12,500 pounds. Are they different categories despite having the same name? If someone can clarify this matter for me and, in case they find the official FAA requirements for the certification of Normal, Utility, and Aerobatic category aircraft, could they please share them with me? I would be immensely grateful

https://www.faa.gov/aircraft/air_cert/design_approvals/small_airplanes/categories
https://www.ecfr.gov/current/title-14/chapter-I/subchapter-C/part-23/subpart-A/section-23.2005

Once upon a time, these commonly-quoted limit load factors were in 14 CFR § 23.337:

(a) The positive limit maneuveringload factor n may not be less than—
(1) 2.1 + (24,000 ÷ (W + 10,000)) for normal and commuter category airplanes,
where W = design maximum takeoff weight, except that n need not be more than 3.8;
(2) 4.4 for utility category airplanes; or
(3) 6.0 for acrobatic category airplanes.
(b) The negative limit maneuvering load factor may not be less than—
(1) 0.4 times the positive load factor for the normal utility and commuter categories; or
(2) 0.5 times the positive load factor
for the acrobatic category.
(c) Maneuvering load factors lower
than those specified in this section
may be used if the airplane has design
features that make it impossible to exceed these values in flight.

This section was removed wholesale when part 23 was revised in 2017. There is no longer such a thing as "utility," "acrobatic," or "commuter" airplanes among newly certificated airplanes. There are only normal category airplanes, and the manufacturer has to show that its limit loads are sufficient for its intended use per § 23.2200(b):

The applicant must determine the structural design envelope, which describes the range and limits of airplane design and operational parameters for which the applicant will show compliance with the requirements of this subpart. The applicant must account for all airplane design and operational parameters that affect structural loads, strength, durability, and aeroelasticity, including [...] (b) Design maneuvering load factors not less than those, which service history shows, may occur within the structural design envelope.

The old categories are still relevant, since the vast majority of general aviation in use were certificated under the old rules and retain their legacy categories.

It should be noted that a means of compliance to show that limit loads are sufficient for the intended purpose may use the earlier g loads defined for earlier category aircraft. The 2017 rewrite moved a lot of regulation text from the FARs to ASTM, which was then allowed to create new means of compliance in parallel with the FAR means. You are not going to be able to design an acrobatic aircraft with a 3g load. You will have to use 6g, some alternate from ASTM or spend a great deal of time proving to the FAA that whatever other number you select is suitable.


text:
Does using a reverse high Cm horizontal tail airfoil coupled with a low Cm wing increase horizontal tail effectiveness?
Does using a high Cm reverse horizontal tail airfoil coupled with a low Cm wing increase horizontal tail effectiveness?
Just noticed that the Zenith 701 wing uses a NACA650-18 which is low $C_m$, while their inverted horizontal stabilizer uses a NACA4412 foil which is a high $C_m$ foil. They could just as easily have used a low cm NACA0018 or NACA2415 foil with a slotted elevator to get the same $C_L$, but they chose a high $C_m$ foil instead
I'm assuming using a low $C_m$ wing and a high $C_m$ horizontal tail adds to the effective moment produced by the area of the horizontal tail, allowing for a smaller horizontal tail area.
Is this correct?

They could just as easily used a low Cm NACA0018 or NACA2415 foil with a slotted elevator to get the same CL, but they chose a high Cm foil instead.

For an elevator, we want to maximize the lift coefficient at the maximum effective control deflection, while maintaining a high degree of linearity with respect to the deflection. There is a maximum effective deflection because any higher control deflection will no longer increase the lift.
A well-designed slotted elevator is theoretically supposed to help with that. But the size of the slot, and any spoiling elements along the hinge would need to be carefully tested in the wind tunnel. Otherwise, they may end up reducing your maximum effective deflection instead. The mechanism required to achieve a good slot may also be complicated for an elevator (slotted flaps don't move like an elevator).

I'm assuming using a low Cm wing and a high Cm horizontal tail adds to the effective moment produced by the area of the horizontal tail, allowing for a smaller horizontal tail area.

That's a good question. I did some quick math using a variation of usual airplane parameters, and here's the summary of the results:

The static margin is unaffected by the camber. For NACA4412, there may even be a forward shift in AC (for the same tail planform), so there may be a slight reduction in static margin.
Having a negatively cambered tail offloads the tail lift, exacting a small saving in trim drag (a few counts at cruise CL of 0.4). Since it offloads the tail, it may also be beneficial to preventing tail stall, but the effect is fairly small.
A negatively cambered tail also produces negative zero-incidence lift. The amount of negative lift can be tailored depending on the overall configuration (CG range, flaps, thrust lines) to minimize form drag.

We can derive the following trimmed lift and pitching moment equations assuming a cambered tail and zero thrust:
(1) $C_{L}=C_{L_{wb_0}}+a\alpha+\frac{S_t}{S}[C_{L_{t_0}}+a_t(i_t-\epsilon_0)]$
(2) $C_m=C_{m_{ac_{wb}}}+C_{m_\alpha}\alpha-V_H[C_{L_{t_0}}+a_t(i_t-\epsilon_0)]+(h-h_{ac_{wb}})C_{L_{wb_0}}+\frac{S_t}{S}C_{m_{ac_t}}=0$
where,

$C_{m_\alpha}=a(h-h_{ac_{wb}})-a_t\overline{V_H}(1-\frac{\partial \epsilon}{\partial \alpha})$
$a=a_{wb}+a_t\frac{S_t}{S}(1-\frac{\partial \epsilon}{\partial \alpha})$
$a_t$ is the lift slope of the tail, $a_{wb}$ is the lift slope of the wingbody, $\overline{V_H}$ is the tail volume from the wingbody AC, $V_H$ is the tail volume from the CG, $\epsilon$ is the downwash on the tail

Perhaps a better description would be a slotted flap which would increase the cL max from 2.2 to about 2.8!!

@Fred tail stall occurs at negative tail incidence angle. Why do you think a slotted flap on tail helps?

Zenith uses a slotted elevator to generate as much downforce as possible, so they can ascend or descend at an AOA of 30 deg for STOL purposes with elevator deflections of 40 deg!!

@Fred Where do you see they use a slotted elevator? On their design page, I only see slotted LE on the wing.

See http://www.zenithair.com/stolch801/design/design.html. for their venturi/slotted elevator, near zero drag slats, low speed flaps, etc.

@Fred I think you mean an elevator where the hinge line allows a certain air gap (that's not a slat)? Most powered planes have sealed hinge lines to improve elevator authority limits. By one account, on a particular airplane model, the air gap (unsealed) seems to have helped. But I don't think this is a universal phenomenon and is model/profile dependent.

I'm pretty sure this is a slotted flap. The max cL of a slotted flap is 1.5 cL base plus 1.3 cL for the slotted flap, for a total of 2.8 max cL, which is exactly the number Zenith is giving.

@Fred Your OP mentioned "slatted elevator". I think you meant "slotted elevator". I've submitted a change to your OP to reflect this.


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What are the maximum possible stabilizer and elevator deflections for the A320?
For research purposes I would like to know the maximum deflection angles (in both positive and negative direction) of the A320-200's stabilizer and elevator.
I am trying to analyse what flight conditions are trimmable.  The outputs I get are deflection angles.  I need to know whether those angles are within the ranges of maximum deflection.

As per the A320 flight manual (§127.20 p3), the maximum permitted total pitch alteration in normal law depends upon speed, aircraft mass, CoG, and other factors:

Pitch Attitude Limitation:
Pitch attitude is limited to:

30º nose up in conf 0 to 3 (progressively reduced to 25º at low
speed)
25º nose up in conf FULL (progressively reduced to 20º at low speed)
15º nose down (indicated by green "=" symbols on the PFD's pitch scale)

The flight director bars disappear from the PFD when the pitch angle exceeds 25º up or 13º down. They return to the display when the pitch angle returns to the region 22º up to 10º down.

Likewise, the maximum physical elevator deflection is 30º nose up and 17º nose down. The maximum trimmable horizontal stabiliser deflection is separately 13.5º nose up to 4º nose down, but the plane will always fly within the total pitch limits specified above.
Good luck with your project!


text:
Do aircraft fitted with both a trimmable horizontal stabilizer and a trim tab exist?
This question is restricted to trim on the pitch axis.
Both trim tab and THS (trimmable horizontal stabiliser, a whole moving horizontal stabilizer whose movements are used as trim) are used to trim an aircraft. Aircraft having a larger speed range (e.g. jet airliners) tend to be fitted with THS whereas aircraft flying slowlier (e.g. GA aircrafts) tend to be fitted with trim tab.
Is there any aircraft type that use both? (Maybe one being a backup device)

The MD-80 has a THS and uses control tabs on the elevators, but not trim tabs. Are you asking only about trim tabs?

Many Messerschmitt aircraft had both, a trimmable stabiliser and a trim tab on the elevator. The tab on the 109 was simply a strip of metal at the trailing edge which could be bent on the ground for adjustment. On the 262 it was a "real" tab.

@Bianfable yes, only trim tabs (I didn't even know control tabs exists, thanks for the info)

@PeterKämpf you should transform your comment in an answer (and I'm going to search for document about Me262 controls, you just awoke my curiosity)

@ManuH: More research revealed that the elevator tab on the 262 is a servo tab (Flettner tab) to reduce stick forces. Trim was done with the stabilizer. Zeno's Warbirds has a flight manual from 1946.

Trimmable horizontal stabilizer (THS) and trim tab serve the similar purpose for two entirely different control architectures:

Trim tab neutralizes the force on the pitch control for a reversible control system.
THS neutralizes the pitch position (and, in the process, the force) for an irreversible control system (e.g. hydraulically/electromechanically actuated, analog or digitally commanded).

In the presence of an irreversible actuator, the actual hinge moment on the elevator no longer matters since the elevator position is being closed-loop controlled by the actuator (provided the actuator is sized correctly, of course); but more importantly, it is no longer felt directly by the pilot. Without this direct feel, a trim tab would be more than useless.
On the other hand, in the case of reversible control with direct elevator-to-stick linkage, deflecting the stabilizer would not allow the dual objective of maintaining trim and neutralizing hinge moment to be accomplished, since a TE up stabilizer would impart even more nose down stick force to the stick. The THS is equally useless for a reversible control for this purpose.
For this reason, I would be very surprised to see an aircraft fitted with both. But maybe there are oddballs?

I've been around light airplanes for about 45 years and I've never seen a trimable stab and trimable elevator using a tab together.  The closest I can come to that is the elevator trim bungee that is included with the jackscrew stab trim on the Piper Super Cub.
The Super Cub has a moveable stabilizer that is the primary trim device, but also has a spring bungee, linking the elevator to the stab, that wants to keep the elevator in line with the stab, instead of just trailing.  This gives the elevator some influence over stick free trim (you could say that it tries to help the stabilizer), improving overall trim authority over a movable stab alone.
Look at Super Cubs on the ground and you'll see that the elevator doesn't hang all the way down, but sits just below neutral.  That's the bungee holding it up.

It's unlikely that any airplane with a THS also has trim tabs.  The benefit of adding them is slight (partial protection against failure of hydraulics), while the cost is extra mechanical complexity and extra drag.  The extra drag is particularly important for airliners.


text:
Was the speed trim system on the B737 Classic modified to counter increased thrust?
How much additional thrust did the 737 classic generate compared to the original 737? Was the speed trim system added to counter the additional thrust?

Wiki says the 1st generation 737s had between 62 and 73 kN of thrust per engine and the 2nd generation ("Classic") between 89 and 105 kN per engine.

Related: When was the Speed Trim System implemented on the Boeing 737?


text:
Why does air density increase with temperature?
Why air density increases with increase in temperature , but this is opposite in case of liquid.

What makes you think it does? Have a look at Density of air.

Density decreases as temperature increases. "Density altitude" increases as temperature increases because the altitude where you find the currently less dense air is would normally be found at higher altitude. The density altitude is just a reference height in comparison to standard conditions

Down voted because the question makes a false assumption.

Maybe the OP is confused with Boyle's law,
a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature. In the case of aircraft flying in the Earth's atmosphere the volume is not constrained.

@Transistor   Yes you are right, I read this line in book on Avionics where they explained concept of duct propagation in radio waves using above mentioned air temp and density relation.

As clarified in the comments, you are confused with Boyle's law, a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
In the case of aircraft flying in the Earth's atmosphere the volume is not constrained so pressure does not have to increase with temperature.

Your premise is false. Air density decreases with increasing temperature.


text:
Can a pilot use glasses to meet the FAA required 20/20 vision?
When the FAA said that pilots require 20/20 vision, does that mean that you have to achieve 20/20 vision with glasses or without glasses?

For what level medical?

Any lvl 1/2 or 3

It can be with correction. And note that for third class, distant vision only has to be correctable to 20/40. The near vision standard is 20/40 for all classes. See https://www.faa.gov/about/office_org/headquarters_offices/avs/offices/aam/ame/guide/standards/
As @Pondlife notes, the operative words here are,  "with or without correction."

So can I be a pilot if I have a power of 20/400 which can be corrected to 20/20 using glasses?

@JohnsonPaul I have around that range of vision in one eye (-5.75 correction) and had no issues getting a Class 2 medical in the UK. I realise yours is FAA centric but I find it hard to believe it'll differ much. Ironically mine corrects to 20/16 (6/5) which is better than average

I'm no AME, but I don't see any reason why not. There's no uncorrected vision requirement in what I linked, anyway.

You might want to quote the most relevant words from the AME guidance: "with or without correction".

And if you need corrective lenses, they become a restriction on your  medical certificate.  Mine states: Pilot must wear corrective lenses for distance vision while exercising the privileges of this certificate. Pilot must have in his possession corrective lenses for near vision when exercising the privileges of this certificate.

Of course, what @Gerry wrote above is very similar to what you'd probably get on a plain driver's license in a similar situation, too. I'm pretty sure this is EU-wide that there is no requirement on uncorrected vision there either, but if you need correction to meet the vision requirements, then a driver's license will also state that correction is required.

One thing to be vary of is laser correction. Do check the regulations carefully before deciding on an operation.

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text:
When was trim on airplanes used for the first time?
I would like to know when trim control was set for the first time on airplanes.. thank you!

Although Anton Flettner patented the trim tab in 1918, (as said in a comment above) it seems that 1910 was the first implementation of "trimming" (according to the following article, linked below)
https://www.flightglobal.com/pdfarchive/view/1910/1910%20-%200255.html
Although it was not adjustable in flight, it still seems like the first time any concept of trim was introduced was in early 1910. However, Anton Flettner produced the first ADJUSTABLE trim tab.

It would be interesting to see if any published  paper airplanes had trim before 1910.

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text:
How does the Airbus auto takeoff system work?
Can you explain in a easy way, how the auto take off system of Airbus works? What is needed on a modern aircraft to perform the take off without help by a pilot?

This is a quote from Airbus's press release:

Rather than relying on an Instrument Landing System (ILS), the existing ground equipment technology currently used by in-service passenger aircraft in airports around the world where the technology is present, this automatic take-off was enabled by image recognition technology installed directly on the aircraft.

So it sounds like the plane is using the same technology used in autonomous cars, which are able to use computer vision and cameras to navigate traffic.  Except in this case the plane just had to keep center-line and rotate at the appropriate speed.
It is also shown in Airbus's video, ATTOL: Autonomous Taxiing, Take-Off and Landing test flight:

Shame, could at least taxi automatically and chat with ATC.

What, are they planning pilot-less flights now?


text:
What are the conditions for the percentage power vs fuel consumption figures in the engine operator manual?
For the Lycoming O-160-A4A, the operator manual says in Economy Cruise, which is 75% Rated power and 2450 RPM, the engine produces 135 HP and burns 9.5 GPH.
How were these figures determined? Were they in standard conditions at sea level, or on the ground at the factory?
It is true that if you assume a power setting to be constant at 65% power and the aircraft is leaned at peak, the aircraft will consume 9.5 GPH, or is this figure not reliable?

Do you have access to the cruise charts or tables?  A blanket statement like that probably = at sea level on a standard day.

The correct source for burn data is not the Lycoming operator manual, but the aircraft manufacturer's manual.
All modern POH have the consumption data qualified (in tables) relative to altitude and temperature.
When adding a new plane to our fleet, we enter that data in and do a linear regression on the data, so that we have a generalized equation for flight planning purposes.
You need to look at the aircraft documentation, not the engine documentation, because the same engine in different planes will have different performance. The POH will have figures which were established empirically, by the manufacturer's test pilots. Fuel burn tests are conducted at various altitudes, and temperatures.
Please note that to answer your fuel burn questions, it is essential to know the aircraft, configuration, and really, have that specific POH.  Then the questions you pose are readily answered.

What would you do if it was an STC for a Cessna 150 and there were no supplementary burn values available?

The same thing as part of your STC, if it was to be a marketed STC.


text:
How does an aircraft maintain level flight while accelerating?
Let's assume a plane is initially moving at constant speed, level flight with a certain angle of attack. If the plane's thrust is increased, the plane's horizontal speed increases (at one point the drag force catches up and becomes again equal to thrust. The speed becomes constant at that point).
But while the speed is increasing, the lift would increase and would the plane start climbing. How can we keep the plane in horizontal level flight while the speed is increasing? Does increasing the thrust need to be simultaneously complemented with controlling some of the rear lift surfaces to oppose the natural lift increase due to the wing?

Simply put, in the scenario stated, to maintain level flight the appropriate amount of nose down input would be applied.
As the op points out, the increase in speed through the air increases lift, and the AOA (angle of attack) needs to be reduced.  This is normally done with stick (or yoke) input, followed by trim.
On most aircraft the lowering of the nose would be accomplished by lowering the elevator (and/or raising the trim tab) on the rear horizontal surfaces.

Assuming a conventional, stable aircraft, most likely yes (unless the changed thrust moment around the centre of gravity balances the aircraft in trim in the new aerodynamic state).
„Normal“ aircraft are statically stable in pitch, meaning that a disturbance in either speed or angle of attack will lead to a change in aerodynamic forces towards restoring the old state. In your case, adding thrust will increase speed and hence lift, so the aircraft will naturally pitch up to restore its old speed and angle of attack (now in a climb due to added thrust). To prevent this and keep level, angle of attack must be reduced to keep lift equal weight at the higher speed.
Note the stability in angle of attack and speed is a simplistic view - real life might be more complicated, as the aircraft will most certainly enter some form of oscillation before (again, on an aircraft of conventional design) settling into a new state.
Some electronic flight control systems are designed to pronounce this self-restoring tendency to the trimmed speed (e.g. that’s Boeing’s Fly-by-wire philosophy), some other flight control systems are designed to suppress it (e.g. on a Fly-by-wire Airbus).
Hence (and slightly on top of what was actually asked), on Fly-by-wire Boeings and conventional aircraft, the throttles are more or less „up/down“ controls and the yokes are more or less „fast/slow“ controls, while on an Airbus it’s vice versa (and sticks not yokes).

"...adding thrust will increase speed and hence lift, so the aircraft will naturally pitch up to restore its old speed and angle of attack (now in a climb due to added thrust). To prevent this and keep level, angle of attack must be reduced to keep lift equal weight at the higher speed..."Yes: more speed, more lift, nose pitches up. Why is the nose pitching up a natural consequence? Is it because if lift increases also the CW moment due to lift increases, I guess. But why is pitching up an attempt to restore the original speed? In what sense? As the nose pitches up, more drag is created

Greater airspeed produces more force on the horizontal stabilizer, which is set to provide a certain amount of nose-up torque (because the center of mass it in front of the center of lift for stability).

"[...] throttles are more or less „up/down“ controls and the yokes are more or less „fast/slow“ controls" Hence the adage "throttle for altitude, pitch for speed" especially during landing.

Most aircraft fly with constant speed almost all the time anyway. They have their optimum cruising speed, which they deviate from only in very rare occasions.

Trim for straight and level just as you normally would. If you increase power you will need to retrim or apply nose down.

The increased airspeed, including the increased prop wash, also flows across the elevator (conventional aircraft), and for a typical (non-aerobatic) aircraft, the center of lift is a bit in front of the center of mass, and the elevator produces a bit of downforce to compensate, and the increased air speed over the elevator would also result in an aircraft pitching up.
For model aircraft, the prop axle is oriented a bit downwards (to reduce pitch reaction to increased thrust), and also a bit sideways (to reduce prop yaw effects). I don't know if full scale aircraft do this.

Compensating for more lift does not require more down force ... more lift while in stable flight WILL result in a climb, period. The compensation is to reduce angle of attack for the purpose of maintaining the same lift at a higher air speed. The horizontal stabilizer does not "push down" on the aircraft as a whole - it pushes up or down on the tail to control this angle of attack.

Engines do not control speed in an aircraft. Elevators control the speed. Consider a glider (sailplane) which can control its speed quite happily without any engine thrust.
Engines control the rate at which total aircraft energy, kinetic plus potential, is gained or lost.
Looking at the exact wording of your question, 'how to maintain level flight while accelerating' ...
Accelerating means increasing speed. To achieve this you push the stick forward. The increased speed means increased kinetic energy, which has to come initially from potential energy, so there is an immediate loss of height. The increased speed also means increased drag, so an increased rate of loss of energy and so height. To maintain height, that is, level flight, you would increase engine power to match the increased drag, and accelerate sufficiently slowly that the engine would also provide the increased kinetic energy.

Yes, an increase in airspeed creates more lift which requires more down force from the tail to keep the airplane from climbing.  To maintain straight and level flight the total lift forces must equal total weight (1G).  In order to compensate for the increased lift more down force must be generated to balance out the increased lift.  So cruising at high speed puts more stress on the wings and horizontal stabilizer / elevator.

The primary objective of the horizontal tail is not to provide downforce to counter lift, but to provide local up/down force in order to modify or balance aircraft pitching moment.


text:
What factors are included in the moment coefficient of a wing?
I am trying to make sure I correctly understand the moment coefficient. Both the lift and the drag coefficients vary with angle of attack and can be either positive, negative or zero. The moment coefficient pertains to the moment specifically due to the aerodynamics force (lift force on the wing mostly). The moment of a force can be calculated about any arbitrary point on the chord (or even outside of it). For a positive cambered wing, if the pole was the arbitrary point X, as the angle of attack varies (positive), the moment would vary and be negative.Is that correct? If the angle of attack goes negative, can the moment coefficient go positive?
If the pole X is chosen to be the aerodynamic center AC (25% of the chord distance from leading edge), the moment of the lift force calculated about the AC remains instead constant with varying the angle of attack.
For a positive cambered wing, I understand that there is a moment due to the lift force (mentioned above) and also another intrinsic nose-down moment (which I think it is constant for all angles of attack) solely due to the cambered shape of the wing. This intrinsic moment is nonzero when the lift is nonzero or even zero. Does the value of the moment coefficient take this intrinsic moment into account as well besides the moment due to the lift force?
Thanks.
Brett

As much as I try to explain aerodynamics without resorting to potential theory, sometimes it is indeed helpful. If you interpret the pressure distribution around a wing as the sum of

a zero angle of attack distribution due to camber and
the Birnbaum pressure distribution due to angle of attack,

things should become clearer.
On the left I show a cambered airfoil at zero angle of attack and on the right an uncambered airfoil at different angles of attack. The first cambered airfoil is a Joukowsky-type which has an arc section for its camber line. The one below it has an irregular camber line that is simply the center between its two surfaces. If you draw circles inside the airfoil at different chord locations, the camber line is the connection of the centers of those circles. The higher the local curvature of the camber line, the higher the local lift at that chord station.

The remarkable thing about the uncambered airfoil is that its lift distribution over chord is geometrically similar between different angles of attack. It only scales in magnitude, but keeps its general shape. Therefore, its center of pressure is at the quarter point for all angles.
The lift due to camber does not change with angle of attack and can be added to the lift of the symmetric airfoil in order to arrive at the total lift for a cambered airfoil at a nonzero angle of attack. Since the lift due to camber dominates the total lift at small angles of attack and becomes more and more insignificant as the angle of attack increases, the center of pressure will move from the point due to camber to the quarter point as the angle of attack rises.
If the lift is composed of a force acting at the quarter point plus an offset moment, this moment is only determined by the camber lift. Since that is constant over angle of attack, so is the offset moment.
This is all strictly true only in inviscid 2D flow, but gives a good approximation for real wings with enough aspect ratio and no flow separation. Viscosity will muddle the picture a little and with smaller aspect ratio the center of pressure of the symmetric airfoil will move forward, but for most airplanes the above captures reality rather well.

For a positive cambered wing, I understand that there is a moment due to the lift force (mentioned above) and also another intrinsic nose-down moment (which I think it is constant for all angles of attack) solely due to the cambered shape of the wing.

Only if you decide to let your lift force attack outside of the quarter point. Then the lift moment caused by the distance between the force vector and the quarter point will produce that moment due to the lift force. If, however, the lift attacks in the quarter point, only the intrinsic nose-down moment which is constant for all angles of attack will be left.

This intrinsic moment is nonzero when the lift is nonzero or even zero.

Yes, indeed. At the right negative angle of attack, the downforce from that angle of attack will exactly compensate the camber lift, leaving the whole airfoil at zero lift but still with the constant moment.

Does the value of the moment coefficient take this intrinsic moment into account as well besides the moment due to the lift force?

Again, only if you (against all convention) decide to define your lift force in a point different from the quarter point. As the moment coefficient is defined, it contains only the "intrinsic moment".
When separation starts (and even with a thicker boundary layer) this will gradually break down and the moment will fluctuate. But then we move outside of the linear range of aerodynamics and potential theory will no longer apply.

Thank you Peter! In essence, when the total lift force is applied at the quarter point, the moment coefficient only captures the intrinsic moment due to camber. This coefficient is negative. You mention that "The lift due to camber does not change with angle of attack and can be added to the lift of the symmetric airfoil". The two figures on the left (positive camber and arcshaped camber) show different circulation distributions for zero angle. So those circulations (I guess lift comes from circulation?) would be the same no matter the angle of attack, correct?

@BrettCooper: What you write in the comment is correct. Lift is roughly circulation times chord. Circulation is another word for vortex strength.

Hi Peter, is this constant aerodynamic moment solely due to the wing camber a pure couple moment (two equal and opposite non collinear forces), that is, a free vector that can be applied anywhere and does not depend on the reference point?

@BrettCooper: Yes, the constant moment is purely from camber, but is produced by a multitude of opposite forces (which result from the integration of pressure over a surface). There is no distinct pair of forces unless you choose to express the result of the camber-induced pressure distribution as those two forces. And yes, this moment is a free vector which does not depend on a reference point.


text:
How are NOTAMs created so consistently?
In the US N.A.S., the NOTAM system is a constant stream of notices getting created, updated, and expired, on topics of airports (big and small), navaids, airspaces, and more.  They frequently use awkward terminology and obscure acronyms.
Yet I cannot remember the last time I saw a typo or mistake in a NOTAM.
This leads me to believe they are not just manually typed in by individual creators across the nation.
What kind of process is there for the creation of NOTAMs that provides this extreme consistency and high quality, and total lack of typos or mistakes?

(the "worst" NOTAM mistake I ever found was a TFR that was centered on a VOR, but the VOR had been shutdown about a year earlier.  As soon as I brought it to the attention of the local office, it was updated.)

Most NOTAMs at commercial airports will be submitted by Airport Operations agents/officers. These are a team of airport employees responsible for ensuring the airport is in compliance with FAR PART 139. They are trained specifically for the task of submitting NOTAMs. This team performs continuous inspections of the airfield and will submit a NOTAM for any item/condition out of compliance; lights not working, unreadable or incorrect signage, obscured surface markings. They are also aware of all construction and planned construction projects and will submit NOTAMs prior to the start of each project.
This is a frequent task for Airport Ops, so they are trained and well prepared by their superiors. However, despite their careful attention, they are human and still make mistakes on occasion. Each submitted NOTAM is then reviewed by Flight Services before they are published. But they too are human, unless they have automation that I am unaware of, and can still overlook a typo.

As long as the database isn't down NOTAM's are managed centrally and its the job of flight service to make sure they are correct. You dont just submit it willy nilly and it gets published. To file a NOTAM you either call flight services or submit it online. If you call the briefer to to file a NOTAM its their job to make sure its correct, if you file it online the system presumably sanitizes it and makes sure it is correct.

It sounds like you're describing a PIREP rather than a NOTAM?  I guess there are a few NOTAMs you can file with a briefer, but that's not the typical way to report Runway issues or closures or TFRs or other common NOTAM issues.

@abelenky Pilots can indeed file NOTAMS (see the link) as can airport managers and other individuals that need to report relevant information.


text:
What is Effective Wingspan?
what is Effective wingspan and why it is said ground effect increase "effective wingspan" of an aircraft? Why would vortex affected by surface result in an increase of "effective wingspan" and lift generation?

As you see on your figure, you get airlift on the entire "effective wing span", so larger surface. So you get some extra "virtual wing".

Related, with Peter's doubt that "effective wingspan" works:  https://aviation.stackexchange.com/questions/35069/how-do-hoerner-wingtips-work


text:
What is the theoretical justification for "proportional navigation (PN)" guidance law?
In missile guidance theory, one of the fundamental guidance algorithms is the so-called "Proportional navigation", which directs a constant-speed missile into a collision course with the target by continously turning the missile velocity vector at a rate proprtional to the rate of rotation of the line of sight to the target and at the same direction (for non-planar scenerios, the last mentioned fact has non trivial importance).
For slow targets (when compared to the missile) and low angle engagements, the closing speed $V_c$ of the missile is constant in time and approximately equal to the missile speed. Therefore, the proportional navigation rule can be stated as:
$$\vec{a} = -N \vec{V_c}\times \vec{\dot\theta}$$
where $N$ is a contant and $\vec{\dot\theta}$ is the angular velocity vector of the line of sight. The logic behind this law is that the greater the rate of rotation of line of sight, the greater acceleration is needed to be applied to set the missile into collision course with the target. If the line of sight is constant, than the missile and the target are already on collision course, and no acceleration is needed to be applied.
However, I read that there are deeper reasons for this law. For non-manuvering target, PN guidance law with $N=3$ minimizes the so called "total control effort" of the missile, a kind of criterion that quantifies the amount of manuvering required for the missile to intercept its target. The control effort is defined as:
$$\int_0^{t_f}|a(t)|^2dt$$
To see why the last statement is right, let us describe the motion of the target from the reference frame of the missile in a planar engagement. Under the assumptions of slow target and low angles engagement, we can write the motion of the target in polar coordinates system $(r,\theta)$ centered at the missile. The acceleration in polar coordinates is:
$$\vec{a}=(\ddot r-r\dot\theta^2)\hat{r}+(2\dot r\dot\theta + r\ddot\theta)\hat{\theta}$$.
Under the previous assumptions we can suppose $\dot r = -V_c, \ddot r=0$. In addition, since all acceleration is applied perpendiculary to the closing velocity vector, we can derive the differential equation:
$$N V_c\dot \theta \hat{\theta} =(2V_c \dot\theta+V_c(t_f-t)\ddot \theta)\hat{\theta}$$
, and if we denote $z=\dot\theta$ than we get the RDE:
$$\dot z = (\frac{N-2}{t_f-t})z$$
whose solution one can get by seperation of variables, which is :
$$\dot\theta (t)=\dot\theta (0)(\frac{t_f-t}{t_f})^{N-2}$$
Now, to connect this result with the variational problem of minimum control effort, one has to impose certain constraints on the small variations of the control function $a(t)$ - the requirement for interception at time $t_f= \frac{r_0}{V_c}$ is equivalent to the the following condition on the variation $\delta a(t)$:
$$\int_0^{t_f}\delta a (t_f-t)dt=0$$
The logic behind this condition is that for interception, small accelerations have to be weighted by the time-to-go $t_{go}=t_f-t$, since earlier velocity changes have greater effect on miss distance.
Now, if we set $N=3$ we get:
$$\dot\theta(t)=\dot\theta (0)(\frac{t_f-t}{t_f})\implies a(t)=3V_cC(t_f-t)$$
where $C$ is a constant.
Now, since the extermal principle requires:
$$\delta \int_0^{t_f}|a(t)|^2dt = 0\implies \int_0^{t_f}2|a(t)|\delta a dt  = 0$$
Plugging inside the last requirement the acceleration formula found before, one can see that for every small variation consistent with the constraint mentioned before, the acceleration function sets the control effort integral in an extremum.
Questions
All this long discussion was just intended to illustrate the connection between PN guidance law and optimization of total control effort. However, I have a much more basic misunderstanding: why the quantity that captures the notion of "amount of manuvers" is defined to be:
$$\int_0^{t_f}|a(t)|^2dt$$?
Why not, for example, $\int_0^{t_f}|a(t)|dt$?

Because this is the only special case that a analytical solution exists? You can arbitrarily a metric function but the majority of them do not derive to anything useful.

I do not think that the total control effort is fixed to be the integral of squares. For example in this technical report, the total control effort seems to be defined as the sum of lateral accelerations in a time-discrete setting.
In the context of optimization, the cost function is often squared as it offers a number of advantages:

The cost function is smooth, which often makes it possible to analytically derive solutions
It discourages large values, which is advantageous as real-life constrains (e.g. lateral acceleration constraints) are less likely to be violated.

A good example (in my oppinion) is the Linear-quadratic Regulator, which features a quadratic cost function resulting in a very nice, linear feedback law.
For these reasons, I do not think that the total control effort is universally defined to be
$$\int_0^{t_f}|a(t)|^2dt$$
but rather often is defined to be in this form to serve as a optimization cost function or serve other purpose in the framework of the paper.


text:
Do the 787 Flight Spoilers have variable deployment when using the Speedbrake?
On the Boeing 787 (other Boeing models welcome too) while in-flight and/or on the ground do the SpeedBrakes/Spoilers have a variable amount of deployment (in angle) between the "ARMED" and "UP" settings? If you have a reference I'd be interested.

You are correct I meant 787. Title updated.

As stated in the answer "The angle of the spoilers varies continuously as the speedbrake lever is pulled aft. AFAIK, this is true for every Boeing model (and also Airbus)."

The angle of the spoilers varies continuously as the speedbrake lever is pulled aft. AFAIK, this is true for every Boeing model (and also Airbus).
On the Boeing 737 NG the actual angles vary as follows (note that ground spoilers only extend on ground, as the name suggests):

Speedbrake lever angle
Detent
Flight Spoiler Position
Ground Spoiler Position

(Nos. 2,3,10&11 / 4,5,8&9)
(Nos. 6&7 / 1&12)

0
DOWN
0
0

4
ARMED
0
0

5-8
-
Start to move
0

29
-
Between 0 and 20/22.5
52 / 60

35.5
FLIGHT
19.5 / 24.5
52 / 60

48
UP (no detent)
33 / 38
52 / 60

(b737.org.uk - Flight Controls - Spoilers / Speedbrakes)

Yes, the farther back the handle, the more extended they are. The slower you pull or push the handle, the more gradual the change in drag. Stow them slowly, it's a very smooth process. Stow or deploy them abruptly, you'll get a pretty noticeable bounce.
No source I can link to, but I do fly 737s, if that's worth anything.


text:
How would a GE9X's react to ingesting a person?
Was wondering how a GE9X (or GE90/etc.) would fare after ingesting a person, at takeoff or climb thrust. (For a story.) Obviously the person is dead and the engine can safely be assumed to be a write off. The question whether a contained failure is likely, unlikely or impossible in that case, (only potentially possible at all due to the size of the engine), and if uncontained whether it would be more likely to throw a fan blade or three, the fan rotor break up, half the compressor going with it or outright desinagration.
Thanks.

Haven't we had a similar question before?

I looked for it, and couldn't find anything here.

It wouldn't react well since the GE9X is vegetarian

Jokes aside this answers should help

Most likely with indifference after a slight cough.

I think it would depend significantly on the forward velocity of the aircraft at the time. Taking bird strikes as an analogy, the bird imparts a very high force to the blade over a short period of time. In engineering terms, we refer to this as an impulse. Roughly speaking impulse = average force * duration of interaction.  The impulse is equal to the imparted momentum, where momentum = mass * velocity.
Assume the aircraft is in forward motion at the time of the impact, and assume the impact takes place near the tip of the fan blade. Then there would be two components of the impulse. A tangential component due to the rotational velocity of the blades, and an axial component due to the forward velocity of the aircraft.
Although I can't quantify it, I believe the tangential component would be relatively low. The metal leading edge of the blade acts as a fairly efficient knife and cuts through ingested objects without too much force.  I've seen high speed videos of bird strike tests.  The fan blades cut through the bird like a hot knife through butter. The blades just slice up the bird and keep on going as if nothing happened.
The axial component of the interaction though, is absorbed by the fan blades by the portion of the bird that hits them head on.  This causes the blades to bend back, and if it is too large could cause them to break.
FAR 33.76 requires the engine manufacturer to test the ingestion of a large (8 lb / 3.65 kg) bird at a forward velocity of 200 knots.
So if a person of, say, 20 times larger mass (160 lb / 73 kg), was ingested near the beginning of the takeoff roll at a 20 times lower velocity of 10 knots, then since the total momentum is about the same, I think the result would be mostly the same as the results of a 33.76 bird ingestion.  That is to say, no hazardous engine effects (i.e. no uncontained failures), although a loss of thrust might be experienced.
On the other hand, if a person of 20 times larger mass was ingested near takeoff rotate at a forward velocity of 130 - 160 knots, then the axial momentum would be an order of magnitude higher than the engine OEM tested during engine certification. There is no way to tell for sure what would happen in this case (at least not without contacting the OEM and asking them to analyze the situation with finite element models).  However, failure of one or two fan blades is probably not unreasonable.  I would expect that such a failure would be a contained failure though, as the 9X has passed the 33.94 blade containment test.  For purposes of a fictional story, if you stretched it to three blades and said it was an uncontained failure, that would still be within the world of believability (but again, only if the impact occurs at  high forward velocity)


text:
NASA's and Boeing truss-braced wing fuel savings breakdown
According to its Administrator Bill Nelson, NASA is working on a project with Boeing that will reduce single-isle aircraft fuel consumption by 30%. He described it as a higher and thinner wing and said...

will reduce fuel consumption by 30%.

and

you can get 30% combination between the engine and the wing.

I think that he's may be referring to the Boeing Truss-Braced Wing project.
My question is what portion of the anticipated fuel savings is directly attributable to the new wing design and what to the engine? Does the engine's contribution depend on the new wing design or will it happen regardless? Is the 30% reduction actually a realistic engineering expectation?
(Note: I've read other posts on this stackexchange about the same project but none of them answer my question.)
Update
I found a Jan 18th, 2023 NASA article about The Sustainable Flight Demonstrator. It said

NASA’s goal is that the technology flown on the demonstrator aircraft, when combined with other advancements in propulsion systems, materials, and systems architecture, would result in fuel consumption and emissions reductions of up to 30% relative to today’s most efficient single-aisle aircraft, depending on the mission.

This should help.

No, those answers don't talk about fuel efficiency other than referencing an aspiration of Boeing's (from 11 years ago) to reduce fuel consumption by 70% using multiple technologies. This answer and the subsequent discussion between sophit and @PeterKämpf is closer to the kind of answer I am hoping for.

Perhaps I should have said above, "style of answer I'm looking for". The referenced answer does not answer my question.

I streamlined the question and voted to reopen it

Thanks. Good edits!

Thanks. Don't know if the question is answerable though, I suppose this information is confidential. Let's wait and see.

There is no defined aircraft or powerplant in this question, so any answer would be speculation. Specifying one characteristic to change and asking about potential efficiency improvements might work, ie. 737 swept wingspan of 117ft changed to straight wing of 170ft.

My hope is that somewhere there's a good explanation of the numbers that Bill Nelson stated to congress. One generally does not just make stuff up when testifying before a congressional committee.


text:
What is the distance from an airport from which an airliner climbs above 550 m?
Is it possible to determine a common distance from the runway threshold at which an airliner climbs above 300 metres and 550 metres above-ground-level (AGL)?
And for landings? Where do they descend below it?
I need to determine at least an approximate value of common distance. But in  the best case I’d like to determine:

the maximum distance
the minimum distance
the average distance or the distance which is used by most airliners.

I am not from aviation, thus I need to ask this question to professional pilots.
The aim is related with the “Bird strike problem”.
300 m and 550 m – is respectively the common height and the maximum height of flight of Black vulture - a bird species living in South and North Americas which caused the most strikes with aviation within its range. I’m doing the scientific research and study the risk of collision with this bird and the possible measures to decrease this risk.

oh, come on guys, you and this "not clear" votes. read the question and is clear plenty, needs only a bit of polishing due to unfamiliarity of the OP with the terminology.

I do not understand the question.

@user6035379 "When are planes between 300m and 550m in regards to the runways (specifically horizontal distance)?"

@user6035379 - "It is better to remain silent and be thought a fool than to speak and remove all doubt" - variously ascribed to A. Lincoln, M. Twain, A. Burns, the Bible (Proverbs 17:28), W. Shakespeare, J. M. Keynes, Empeco, L. F. Baum, St. Francis de Sales, Doc Rockwell, Mother Goose, Maurice Switzer, Confucius, and me.

You may also try tracking websites, e.g. London Heathrow. Hovering an aircraft icon with the mouse shows its height. Other sites: FlightRadar24 -- FlightAware -- many other ones. These sites let you play with archived flights, you don't need to wait for actual departures/landings. You may also ask them whether they can and want to draw the contours on a map automatically from their flights database.

For the landing is quite easy.
Airliners usually approach an airport descending along $3^\circ$ slopes. This means that they will be 300m AGL approximately 5.7km before the runway threshold, and they will be at 550m AGL about 10.5km before the same point.
For the take-off the situation is a bit more complicated, as each aircraft will have a different performance. Taking a value of about 1000ft/min for the vertical velocity and 200Kts for the forward velocity (in aviation we use the imperial system, for a bit of background see here) we get similar distances to the above (it is a $2.8^\circ$ slope)
Note that these paths are more or less a straight continuation of the runway, particularly on the landing side (less so on the take off side).
Each airport has approach and departure procedures, you could try to have a look at those. They are usually published by Jeppesen.

It might be useful to add that these paths are aligned (more or less) with the runways.  A bird 6km from the runway at 350m is less at risk of collision if it's not on an approach or departure path.

@DanPichelman you're right, I'll edit.

I think a figure closer to 1500-2000 fpm would be more accurate for climb.

@JonathanWalters I'm open for improvements. I couldn't find good data on the spot and I went for a conservative figure.

3° is the typical standard, but there are exceptions.  London City Airport, for example, flies a 5.5° approach glideslope (because of noise abatement rules).

Ground distance for takeoff will also be affected by headwinds, and factors such as passenger load & local noise-abatement rules.  IIRC, departure from Orange County required a steep initial climb, then further climbout at reduced power...

@J... Sounds like my kind of an approach!

@J... : While the City has its share of vultures, those tend to hang out in offices, not at 1000 ft AGL.

There are no pre-set figure for this, as it depends on the airport and surrounding geography, aircraft performance, departure and approach procedures. ATC instructions, etc.
For departures, airports don't publish a standard point where you must reach these altitudes but they sometime publish Obstacle Departure Procedures (ODP)  which will specify a minimum climb gradient which the performance capabilities of your aircraft must be able to achieve in order to safely fly the ODP.  This gradient is usually listed in hundreds of feet per nautical mile.  ODPs assume that you, the pilot will be able to achieve an initial climb of 500 FPNM and an altitude of at least 35 ft above field elevation by the time you clear the departure end of the runway, climb straight ahead to an altitude of at least 400 ft AFE prior to beginning your first turns from runway heading, thence maintain a climb gradient at or above the minimum listed to the waypoint listed on the ODP.
An example is as follows: Glenview Airport (KABC) has a RIVER1 ODP, which requires your aircraft to depart runway 27 thence
"climbing right turn to a heading 307° to intercept the 287° radial from the ABC vortac to CYRAC intersection 16 DME from ABC at or above 4500 feet, thence resume own navigation.  Minimum climb gradient of 450 feet per NM to CYRAC."
If you are flying a C-172 and assuming a Vy of 76 KIAS, you must achieve a rate of climb of (450*76)/60 = 570 feet per min from the departure end of runway to CYRAC intersection in order to safely fly this departure procedure.
Descent and landing will usually involve the use of feeder routes from established Juliet (Jet) airways onto Standard Terminal Arrival Routes (STARs) which have established step down fixes at specific waypoints along the STAR or as advised by ATC.  The STARs then feed into the specific instrument approaches for the destination airport which also has step down fixes prior to final approach fix which descends from there to either a missed approach point (non-precision app) or a decision altitude (precision app).  A pilot must have confirmed visual acquisition of the runway environment from that point and the weather at the airport must be at or above the published minimums in order to continue to approach and land, else the pilot must go missed and fly the missed approach procedures for that plate.

authors: Bob Jarvis - Слава Україні


text:
Why don't commercial aircraft adopt a slightly more seaplane-like design to allow safer ditching in case of emergency?
For example :

Why don't they have a more hydrodynamically shaped fuselage underside ?
Specifically for the aircraft that have their engines mounted under the wings, isn't this too dangerous for ditching ? As far as I understand when an engine touches water during ditching the whole aircraft would/could break apart ?
Or, couldn't they have some kind of (retractable) floats ?

I'm far from being an aviation expert myself, but I imagine some things like the above (or possibly others) could be done, in order to allow safer ditchings for commercial aircraft. Would all these be too costly for airlines ?

Related: https://aviation.stackexchange.com/questions/9564/why-dont-big-commercial-planes-have-full-aircraft-parachutes

There are more than 100.000 commercial flights/day. How many ditchings per day? See the issue?

Note that this was the case in early aviation when reliability and infrastructure made the B314 economically interesting. Today's, the loss in aerodynamic due to a hull and the costs of such modifications may be significant enough compared to the slight advantage you may gain (note that today's airliners are able to ditch) to not implement it.

There has been another question about whether some commercial aircraft have "breakaway" underwing engines, i.e. a shear pin, or something like that, and why.  https://aviation.stackexchange.com/questions/23151/how-are-engines-mounted-onto-wings , https://aviation.stackexchange.com/questions/24935/have-jet-engines-ever-sheared-off/24940#24940 , An interesting ASE question would be to ask whether this feature is intended to help, or ever actually does help, in a ditching situation.

@Hanky Panky according to your logic there shouldn't be any life jackets on aircraft either (or other similar equipment that would be used in case of landing on water)

Seaplanes are designed to take off from water, ditching an aircraft doesn't require it to take off again, perhaps they already incorporate everything needed to safely ditch but not take off again?

If airliners were easy to ditch, we wouldn't have the Sully movie, would we? See the issue?

@QuoraFeans the problem is the only time you want to ditch is when the aircraft is already in trouble, so yes its going to be difficult cos the aircraft is not operating correctly

See, @scrontch has it. Also, say the aviation industry decides no cost is too much to save lives, right? Then flying becomes more expensive than driving for more trips, and people choose/have to drive, which is way more dangerous than flying, so the net death rate actually increases.

because then passengers would never board.

Because ditching is extremely, extremely rare, so the costs of redesigning aircraft along with the extra drag and weight (increased fuel burn) it would no doubt add to the airframe far outweighs the potential benefits.
That argument might sound weird to someone, but think about it this way: would it seem reasonable to redesign every single car on the planet so that they can float, just because every year a few people around the world accidentally drive their car into a lake?
A quick search on The Aviation Herald suggests that, since 2005, a total of four flights worldwide have been forced to land on water. To give you an idea of the scale, there are about 100,000 commercial flights worldwide daily. So that's something like 4 flights out of half a billion, which is practically nothing. Also, in those 4 accidents, more than half of the people onboard actually survived. Actually, the increased greenhouse gas emissions caused by adding extra weight, such as floats, to aircraft worldwide would probably kill many more people than such safety equipment would ever save.
It's not that the aviation industry isn't investing heavily to improve safety. It's just that making ditching safer is an area where there is almost nothing to be gained. So the money can be much better spent elsewhere.

+1 for the excellent observation that the increased CO2 emissions from the higher weight of ditching-adapted aircraft would kill more people than would be saved in ditchings.

It always amused me that 80% of the pre-flight safety demonstrations the crew are required to give on every flight are all about what to do in the event of a water landing, given how rarely that occurs.  (But that said, there are in fact already precautions for water landing - those safety vests under the seats and the life-raft emergency ramps.)

Also, adapting a plane's structure to be better at landing in water will surely have drawback on its flying characteristics. Given how unlikely water ditchings are, I think we're safer with planes optimized for flying between designated airports than trying to be jack-of-all-trades.

Similarly, one could ask why people about to ride the New York City subway do not first stock up on snake antivenom, or why cattle ranchers in Texas tend to own only small arms (shotguns, rifles, and pistols) rather than mortars, flak cannons, and Surface-to-Air Missile batteries, even though there is a theoretical (nonzero) danger that some hostile military will decide to buzz Farmer Bill's flock.

4 ditches out of half a billion flights... As regular emergency landings happen way more often than ditchings, I would guess the added weight of the floats would cause more than 4 crashes out of half a billion, or turn survivable crash landings into unsurvivable ones.

the increased greenhouse gas emissions caused by adding extra weight, such as floats, to aircraft worldwide would probably kill many more people than such safety equipment would ever save. — doesn't that apply to the weight of life vests too? Or more to the point: I don't believe this is the reason at all.

@gerrit: Even if the airline's argument is financial (fuel cost) rather than ecological; the same point still stands. This answer does not claim that ecology is the explicit reason for the decision made by airlines, only that it is an additional side effect of increased fuel usage.

On top of the previous good answers, I would like to add that the aerospace industry does take ditching seriously. 14CFR Part 25 and regulations from other agencies require that transport category aircraft be designed such that the occupants have a reasonable chance of surviving a water landing. This means that, in addition to providing life rafts and life jackets adequate for all occupants, the structure of the aircraft must allow for water landing and enough flotation time post-impact for evacuation (i.e. not break up, catch fire on impact).
It is up to individual applicant to determine what is the best course to satisfy the laws. If the applicant determines, after engineering/costs trades, that installing water skiis and changing the hull shape are the best way to meet the regulation, then so be it.

This is a very good point: modern airlines can effectively ditch with decent survivability. The seaplane hull shape (and spray-protected engines) are necessary only if the plane is to take off again from water...hardly a concern if you've ditched owing to engine failure.

@Jimmy this may raise a whole new discussion but, since aerospace industry takes ditching seriously and as you say there are related regulations, why isn't ditching part of commercial pilots mandatory training ? IMHO any kind of safety system/ item / aircraft design related to ditching is totally useless if the pilot has no clue how to - at least theoretically - perform an acceptable landing on water.

@funkyjelly: Do you mean real life training? (If you mean simulators, I'd agree but unsure as how effective simulator training may be) Because that's quite dangerous to have a rookie perform in real life. Any failure on their part is going to (at best) rack up repair costs, or (at worst) kill everyone on board. And I'd hazard a guess that ditching an airliner is not the same as ditching a twoseater. Having rookies train for this in actual airliners is going to be insanely expensive and dangerous.

@Flater I fully agree that it's not something that could be trained on real conditions. I'm talking about theoretical and sim training.

@funkyjelly Not a pilot, so can't speak for pilot training. Ditching procedure is in the QRH and non-normal of AFM of transport category aircraft. It's very tough to simulate water landing that is representative, so I'm doubtful of the value of simulator (in their current form) training for ditching.

This answer wins: they already do that.

@funkyjelly, because there is nothing so specific to ditching to require it. The landing is mostly normal except you let it float and settle as gently as possible, but that's learned from normal landings, and if there is time to make a proper emergency landing, there is time to read the checklist, so no need for memory items.

Given the previous answer, I wonder if the costs of those regulations (in terms of additional fuel, for example) do not outweigh the benefits.

Multiple reasons:

Ditchings are extremely rare as it is, so if the goal is to improve safety there are plenty of other areas that can be worked on first, like Crew Resource Management.
It is much easier to improve the reliability of the engines than it is to ditch-proof a high-subsonic airframe.
The structural weight penalty for a seaplane hull is quite large and would directly translate into higher fares.
A hydrodynamic hull is not suitable as a pressure vessel for pressurized cabins because of the sharp angles on the underside acting as stress concentrators. The simplest solution would be to add a pressure hull inside the hydrodynamic hull, further increasing weight.
The aerodynamic penalty for floats is massive at the kind of speeds modern airliners cruise at.
Overwing engines have notable drawbacks due to their inaccessibility for maintenance.

Why don't they have a more hydrodynamically shaped fuselage underside?

Because it wouldn't help emergency ditching much. All that matters is that the hull stays in one piece long enough to protect its occupants, and changing the shape of the hull would more likely compromise its strength than improve it—a tube is excellent shape for strength.

Specifically for the aircraft that have their engines mounted under the wings, isn't this too dangerous for ditching? As far as I understand when an engine touches water during ditching the whole aircraft would/could break apart?

The whole aircraft probably not, since the engine mounts are the weakest points. If the fuselage does not break in two, anything else that shears off is actually absorbing the impact energy and therefore rather beneficial for the occupants. And the aircraft is going to be written off either way.
The engines are also very dense and heavy, so applying the braking force to them actually produces less stress in the structure then applying it elsewhere.

Or, couldn't they have some kind of (retractable) floats?

It wouldn't help either. If the water is calm enough, the fuselage is strong enough to withstand the impact and deceleration, and seaplanes can't land in much of a swell either. In fact in a swell floats make things worse, because they raise the centre of gravity, so the aircraft has greater tendency to roll over forward and it can't survive hitting the next wave at too steep angle.

This answer is underrated.

You'd need to change every aircraft design in existence to be a flying boat. This isn't a minor change, it's a completely different design paradigm, one that doesn't lend itself well to the speeds used by modern aircraft.
Jet powered flying boats were experimented with, they never were very successful (right now only one model is flying, the Russian Beriev Be-200 water bomber

It works, but isn't economical for use as an airliner or commercial cargo aircraft because of the high operating cost per passenger-mile (or ton-mile) (were it to be operated as such, which it isn't).
And that's not something that can be easily changed because of the very nature of flying boats. The hull design doesn't lend itself to pressurisation, meaning you either need to fit an internal pressure cylinder, adding serious weight and reducing available internal volume or you limit yourself to far lower cruise altitude and speed (increasing travel time and fuel consumption).
The hull shape is also not very aerodynamic, reducing performance and increasing fuel burn even more.
For special operations where requirements like the ability to operate where there are no runways override economics this is no problem, in the airliner business it's a death sentence.

Correction: The Be-200 is pressurized. It essentially tacks on some hydrodynamic elements to make it able to land on water, while staying very close to that conventional cylindrical shape.

Most of the special features of a seaplane are about taking off from the water, because of the need to minimize drag.  Almost none of this is useful for an airliner.
When you're ditching, you're probably going to ditch on the ocean, and unless you go all-out on the seaplane design, salt damage alone is going to destroy the airplane.  Design of airliners for ditching is based around the assumption that the plane won't be re-usable:

The fuselage is reasonably hydrodynamic as a side effect of streamlining.  It lacks the "step" of a seaplane, but you only need that to reduce hydrodynamic drag during takeoff.
The engine pylons are designed to be a weak point.  In the event of a water landing, the engines will break off without significant damage to the wings.
Floats are only useful to keep the airplane steady during taxiing, loading, and other surface operations -- on a seaplane, the floats are the first thing to leave the water during takeoff, and the last thing to touch down during landing.  An airliner is designed to float with the wings right at the waterline, to provide the same stabilization effect.

Any airliner designed for over-water operations is required to be able to ditch safely in conditions where a seaplane would be able to land, and remain floating long enough for the passengers to evacuate.  Adopting more "seaplane-like" features wouldn't improve on this, and would reduce performance during normal operations.


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How possible is a successful landing just with 1 wing?
Having in mind this famous F-15 incident :

Would this be possible with any other fighter-jet? Or was there something really specific to F-15's design that allowed this ?
Would this be possible at all with any commercial aircraft ? Specifically for those that have their engine(s) mounted on the rear ? or what about for those that have engines mounted on the wings, but only with a partial wing loss (eg having both engines intact) ?

There is the famous case of a DC-3 that landed with most of the outboard panel of one wing missing -- and the only replacement panel available (in WWII China) was from a DC-2, so it flew for years as a "DC-2 1/2".  Needed some aileron trim, but otherwise was reportedly fine.

the A-10 was designed to fly with a good part of a wing and one of the elevators and vertical stabilisers missing. I don't know if it was ever put to the test, but during Desert Storm some did come back with a lot of holes in them.

Call me pedantic, but F-15 has one wing total. It's called monoplane for a reason (as opposed to biplane with two wings). In that incident, F-15 lost about a third of its wing.

@jwenting I think Kim Campbell probably brought back the most infamously trashed A-10 to a safe home-base landing - smoking, broken, missing too many parts to count, filled with bullet holes, and yet still limped back to base.

This would be possible with other fighter jets that are of similar design to the F-15, i.e. have a low aspect ratio, much of their lift produced close to the centerline, and a lot of control authority. They would also need the other two key ingredients for a landing of this kind: a strong aviator at the controls and a bit of luck.
Specific contributing features of the F-15 were its fuselage's lifting design, side-of-fuselage engine placement, blended wing with strakes, and large control surfaces, common to fighters.
If you look at the planform, it wasn't missing a full 50% of its wing, but only about 20% of its lift-contributing planform area (lift distribution isn't equal across it of course), with the body, the intakes and wing root strakes continuing to provide lift. As a result, the center of lift didn't move as far away from center as it would have for a tube-with-wings design.
In principle this is a more violent version of a runaway aileron scenario, and one has to compensate for roll from the lost wing area. The F-15, being a fighter, has large control surfaces for combat maneuvering, and they proved sufficient for stopping the roll in this case. The loss of lift was not so much of an issue.

The red box is the lost area, the green box the remaining planform, and the dashed line its very approximate center.
It's possible that other fighter aircraft could do something similar. Commercial aircraft, however, almost universally feature high aspect ratios, and with such a wing, the center of lift will move too far away from the design centerline in a similar scenario. Partial wing loss, however, can be compensated for, and there's a lot of incidents where planes landed having lost some part of their wing.

That's not to say this couldn't be screwed up.  Often when a jetliner loses a fraction of its wing lift, it crashes, because the problem "shows up" at lower speeds, i.e. takeoff and approach, when they have too little altitude remaining to learn.

In addition, if it didn't have thrust vectoring engines it wouldn't have made it.

@Joshua the F-15 doesn't have thrust vectoring engines (except one or two experimental conversions, but those weren't involved here).

It is definitely possible to lose an entire wing and still control the plane if the plane is capable of knife-edge flight.  Check out these videos of it being done-- ending in amazingly successful landings -- with radio-controlled planes.  Of course a full scale-plane being flown in this matter would impact the ground with much more energy in relation to the strength of the structure, but these videos suggest it still might be survivable at least with aircraft similar to the ones represented by the radio-controlled models.   Of course, not in any commercial airliner, and the odds of success with this particular technique would surely be slim in a heavy military jet as well.

(second video intentionally set to start at 2:18)
Also, beware the well-known hoax video that purports to show something extremely similar with a full-scale airplane with the logo "KillaThrill" emblazoned on the side.  It is actually a mix of full-scale and RC footage and no wing was actually lost by any plane in that video.
It's interesting to note that if you want to use an alternative technique involving flight in a near-level bank attitude, then full-span ailerons are a disadvantage and may make it impossible.  You want a powerful (wide-chord) aileron near the wingtip generating lots of downward roll torque, leaving the rest of the wing free to make positive lift.

Videos were found by googling "Landing RC plane one wing"-- many more such videos may exist, I didn't look very long.  PS double-posting answer was not intentional, deleted now.

Are the RC planes able to pull this off so gracefully because of their high thrust to weight ratio?

@GittingGud That, and very high rudder authority

There are single winged aircraft too - but I don't see how this answers the question.

Keep in mind, also, that the scaling laws of the universe mean that what works in an RC plane won't necessarily translate to a similar performance in a larger aircraft.

@GittingGud - any RC plane called a "3D" plane can actually hover with nose straight up and zero airspeed, but we aren't seeing that here (not sure why not).  It does not look to me like extreme thrust to weight is being employed, otherwise it would hang on prop with virtually no airspeed.  As far as extreme rudder authority goes, I'd say no more rudder is being used than is needed for knife-edge flight which any full- scale aerobatic plane is capable of.

@GittingGud -Therefore I believe a typical full-scale aerobatic plane could duplicate these maneuvers up to the instant of ground contact.  What happens after that may be another story.  For example, was hard to tell how much load was being put on low tip of horizontal tail during ground contact.

Listen carefully to the motors- they aren't revved up to high power after loss of wing, or as the aircraft approaches the ground.  A more fruitful line of thought may be what is the loading (or the fuselage side loading in knife-edge flight) and how does it compare to full-scale, what are differences between stall speed model and full-scale, minimum airspeed for low-power descending knife-edge flight model and full-scale, how does kinetic energy on impact scale with speed and weight. Rc planes can survive some crazy-looking landings.

Also as far as structure and weight and impact loads go, you can lift up typical rc plane just by holding outboard tip (say outboard quarter) of one wing with a/c in level bank attitude, this would probably snap the wing spar of a full-scale plane.  Likewise you can drop the rc plane onto the ground in a vertical bank     from a half-span above ground and likely do no damage.  So, no doubt there are scaling issues trying to duplicate the videos w/ full-sized a/c, but I think they mostly involve the result of a hard landing in an unusual attitude.

Looks like it's the landing gear not tail that takes the impact- the gear of full-scale plane would be demolished by a similar maneuver but it still might be survivable.

Or at least, MIGHT be demolished.  Some of those gear design look quite strong.

These landings brought to you by the "square cube law", which gives the models a great 1/scale^2 the rudder authority on 1/scale^3 the mass.


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How do you calculate the vertical and horizontal tail volume if you have Vvbar and Vhbar?
I know the general formulas for calculating the horizontal and vertical tail volumes. Unfortunately, I only have Vvbar and Vhbar. I’ve checked through multiple textbooks and I don’t know why I can’t find an expression for Vvbar.
This is what I have.

The 'bar' volumes are the same volumes but taken with respect to the wing-body aerodynamic centre (AC) rather than CG. This is more convenient because this is a constant reference, whereas CG may move.
The difference between them is only the arm: $\overline{c}(h-h_{n_{wb}})$, which leads to your last expression for $V_H$.
For the yaw motion, the aerodynamic centre of the wing is much less relevant(*), and thus the 'plain' $V_V$ (with respect to CG) is commonly used. This is why you can't find an expression for $\overline{V_V}$.
Nevertheless, if we want to relate it to the same wing-body AC as for $V_H$, we can do it. However, in lateral motion the generic reference is the wing span $b$ rather than the MAC $\overline c$ (as you can see from your first two definitions). Thus $\overline c$ is not eliminated, and we have an uglier expression:
$$\overline{V_V} = V_V + \frac{S_{VT}} S \frac {\overline c} b (h-h_{n_{wb}})$$

(*) It could be related to AC of the body alone (which is typically near the very front of the fuselage), but this is also not very sensible: the wing (and the wing-body interaction) does play a major role. I've never seen $\overline{V_V}$ defined like that for airplanes.

Thank you so much. I really appreciate it. I went in circles for a few days.

You're welcome. You can select the green tick beside the answer to mark it accepted.

You need to multiply the cg moment arm ($h-h_{n_{wb}}$) by the MAC then divide by reference span.

Thank you so much. I really appreciate it.


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Why aren't flights continued after losing a tire on rotation?
I have read about a lot of incidents where airplanes lose one of the tires on rotation, yet they cancel the flights and go back for an emergency landing at the departure airport, risking overweight landing or burn fuel for 2 hours until they are within the MLM.
My question is why don't they choose to continue the flight to the destination since they already airborne?
Fixing tire out of station seems much cheaper than burning or dumping fuel to me.

Sorry, a non-native lurker here - could you please explain the meaning of on rotation?

@CopperKettle "on rotation" is the moment when the plane's nose gets up (take off). The plane "rotates" upwards, so to speak, although it's just a few degrees.

Concorde 4590 was brought down by a burst tire on rotation.

Some do continue - https://aviation.stackexchange.com/questions/29663/what-level-of-damage-to-an-aircraft-is-acceptable-to-let-it-depart

Blown tire - can't raise gear, shredded tire will not fit in the bay.    ->
Gear down - much lower maximum speed and lots of additional drag ->
Both of those mean very poor fuel economy ->
Very poor fuel economy means not enough fuel to reach destination
So that's off the table.  Given that, there's no question that it's better to follow the safest course... And you had an extremely high pressure tire just explode, a couple feet away from the airplane's skin.
Especially when that brings you back to a field that is in fact an operating base for that airline, and is likely to have inspection and repair services, and may have a spare airplane.  I could see a Southwest flight out of SJC diverting to nearby OAK, since OAK is a main operating base for Southwest and will have an easier time fixing or replacing the aircraft.

If I remember correctly, Terry once told us that "gear down" needs almost twice as much fuel compared to "gear retracted" (in the 747s he has flown).

'Nother problem .. they are here at an airport with conditions that allowed start / landing .. which can't be automatically assumed for the target airport

"Blown tire - can't raise gear, shredded tire will not fit in the bay" - and might be on fire.

That kind of thing would be a judgment call on the part of the crew (ultimately the capt) and would come down to what is the safest action based on the circumstances, with logistical/convenience considerations being a distant second.
When you have a blown tire you will want to avoid raising the gear, because they can catch fire and you don't want to retract the fire into the fuselage (and your own fire detection system won't tell you this until it's retracted, so you really won't know until it's too late).  On top of that, you may have engine damage, especially if they are tail mounted, and/or flap damage.
So the imperative is to get the hell back down post-haste.  However, the airport is also a consideration.  Most of the time the departure airport has long runways and good emergency services, so the crew will elect to do a turnback even if fuel dumping is necessary.  In other circumstances, say if the departure airport is remote and has short runways with minimal emergency facilities, the crew may elect to divert to the nearest airport with suitable facilities, and will likely do the diversion leaving the gear down.
Individual airlines may have specific policies on this as well, and a capt at one airline may have more discretion on what to do than another.

Sometime tires are retracted because the pilots dont realise they have blown the tire and only be aware of it when the tower inform them of debris on the runway, assuming destination airport is well equipped to handle this sorta of emergecny, still didnt understand why not proceed with the flight? also i didnt find any source confirm that you cannot retract blown gear ..

It's assumed they would know they blew a tire or they wouldn't have to apply "judgement".  Yes sometimes they don't.  On a heavy or corporate a/c with a heavily insulated cabin there may only be a faint bump sound.  There is nothing printed anywhere that says you CAN"T retract with a blown tire; it's just bad airmanship.  The Nationair incident drove it home.

Because an emergency landing is preferred to gambling the airplane and the lives of everyone on board just to save a few bucks.  One of the fundamental rules of professional flying:  Never attempt normal operations or try tricks with a compromised airplane.  You might get away with it once, but sooner or later it will get you and, with a  $300 million airplane and 270 souls aboard, there will be hell to pay.
Blown tires, with the kinds of tire pressures used, are dangerous.  Depending on what caused the tire to fail, there can be additional damage to the aircraft.  And let’s not forget that you will have to land with a failed tire which has serious damage and fire risks associated with it as well.  An emergency cannot be treated casually under any circumstance.  In the wake of the Concorde accident in 2000, flight crews don’t take risks with that.  Your best policy with a tire failure post V1 is 1) get airborne 2) assess the situation and contact ATC for additional help from the ground 3) select the best airfield nearby or within flying distance for a emergency landing 4) dump fuel for the minimum aboard needed to attempt the landing and 5) conduct an emergency landing with crash and fire crews waiting.

The first bit of the second paragraph is true regardless of if the aircraft lands at the departure or destination airport.

The first part of the first paragraph is also true, regardless of if the aircraft lands at the departure or destination airport.

@jvriesem: not quite: when flying between airports instead of circling one, you're farther from a runway if a problem develops and you need to land immediately, regardless of burning off more fuel.


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What was the fastest single engine plane with predominantly wood frame and fabric skin construction?
I've been doing some research and I haven't found a satisfying answer.
The Mosquito was quite fast but it was wood skinned and had two engines. The Bellanca Viking was fabric skinned but used a metal space frame for the main body.
A good candidate might be the Bell XP-77, a light weight interceptor built out of non-strategic materials in WW2. I've seen it quoted at 329 mph. (https://www.militaryfactory.com/aircraft/detail.php?aircraft_id=1248) Turns out the Bell XP-77 is wood skinned with some kind of fabric layer involved somewhere.